If the random variables X and Y have joint density function
otherwise
x y
xy
f x y
0
0 4, 1 5
( , ) 96
a) The marginal distribution of X and Y.
b) E(x), E(y) and E(xy)
c) Conditional distributions of X given Y
d) Conditional distributions of Y given X
e) P(0 ≤ 𝑥 ≤ 1 𝑎𝑛𝑑 1 ≤ 𝑦 ≤ 2�
a) To find the marginal distribution of X and Y, we need to integrate the joint density function over the respective variable.
For the marginal distribution of X:
```
fX(x) = ∫ f(x,y) dy
= ∫ xy dy
= x * (∫ y dy) [from 0 to 4, 1 to 5]
= x * (1/2) * (y^2)[from 0 to 4, 1 to 5]
= x * (1/2) * (16 - 1)
= 7x
```
Therefore, the marginal distribution of X is given by fX(x) = 7x.
For the marginal distribution of Y:
```
fY(y) = ∫ f(x,y) dx
= ∫ xy dx
= y * (∫ x dx) [from 0 to 4, 1 to 5]
= y * (1/2) * (x^2)[from 0 to 4, 1 to 5]
= y * (1/2) * (16 - 1)
= 7.5y
```
Therefore, the marginal distribution of Y is given by fY(y) = 7.5y.
b) To find E(x), E(y), and E(xy), we need to calculate the expected value using the respective marginal distribution.
For E(x):
```
E(x) = ∫ x * fX(x) dx
= ∫ x * (7x) dx
= 7 * (∫ x^2 dx) [from 0 to 4]
= 7 * (1/3) * (x^3) [from 0 to 4]
= 7 * (1/3) * (64)
= 149.33
```
Therefore, E(x) = 149.33.
For E(y):
```
E(y) = ∫ y * fY(y) dy
= ∫ y * (7.5y) dy
= 7.5 * (∫ y^2 dy) [from 1 to 5]
= 7.5 * (1/3) * (y^3) [from 1 to 5]
= 7.5 * (1/3) * (125 - 1)
= 520
```
Therefore, E(y) = 520.
For E(xy):
```
E(xy) = ∫ ∫ xy * f(x,y) dx dy
= ∫ ∫ xy * (xy) dx dy
= ∫ (x^2 * y^2) dx dy [from 0 to 4, 1 to 5]
= (∫ (y^2) dy) * (∫ (x^2) dx) [from 0 to 4, 1 to 5]
= (1/3) * (y^3) * (1/3) * (x^3) [from 0 to 4, 1 to 5]
= (1/3) * (125 - 1) * (1/3) * (64)
= 888.89
```
Therefore, E(xy) = 888.89.
c) The conditional distribution of X given Y can be found using the formula:
```
fX|Y(x|y) = f(x,y) / fY(y)
= (xy) / (7.5y)
= 2/15x, 0 < x < 4 and 1 < y < 5
```
Therefore, the conditional distribution of X given Y is fX|Y(x|y) = (2/15)x.
d) The conditional distribution of Y given X can be found using the formula:
```
fY|X(y|x) = f(x,y) / fX(x)
= (xy) / (7x)
= (y/7), 0 < x < 4 and 1 < y < 5
```
Therefore, the conditional distribution of Y given X is fY|X(y|x) = (y/7).
e) To find P(0 ≤ 𝑥 ≤ 1 𝑎𝑛𝑑 1 ≤ 𝑦 ≤ 2), we need to integrate the joint density function over the given range.
```
P(0 ≤ 𝑥 ≤ 1 𝑎𝑛𝑑 1 ≤ 𝑦 ≤ 2) = ∫∫ f(x,y) dx dy
= ∫∫ (xy) dx dy [from 0 to 1, 1 to 2]
= ∫ (y/2) dx [from 0 to 1] * (y) [from 1 to 2]
= (1/2) * (y^2) [from 1 to 2]
= (1/2) * (2^2 - 1^2)
= (1/2) * (4 - 1)
= 1.5
```
Therefore, P(0 ≤ 𝑥 ≤ 1 𝑎𝑛𝑑 1 ≤ 𝑦 ≤ 2) = 1.5.